I am using Spivak Calculus on Manifolds to understand differential form. I understand that a k-form is defined to be the function sending a point to this point paired with an alternating k-tensor. I have confusion of the following discussion:
If $f :\mathbb R^n \to \mathbb R$ is differentiable, then $Df(p) \in \Lambda^1 (\mathbb R^n)$. By a minor modification we therefore obtain a 1-form $df$, defined by $$ df(p)(v_p) = Df(p) (v).$$
I wonder what does $Df(p)$ means here. Why does the derivative of $f$ at a point $p$ become a 1-tensor instead of a point in $\mathbb{R^n}$? Does the RHS mean $Df(p)\cdot (v)$?
Perhaps a little late to the party, but if you read through Chapter $2$ again, Spivak defined the derivative of a function at a point to be a linear transformation. So, if $f$ is a function from $\mathbb{R^n} \to \mathbb{R}$, and if $f$ is differentiable at a point $p \in \mathbb{R^n}$, $Df(p)$ was defined to be a linear transformation $Df(p): \mathbb{R^n} \to \mathbb{R}$. So, $Df(p) \in \mathcal{L}(\mathbb{R^n}, \mathbb{R}) = \Lambda^1(\mathbb{R^n})$. Hence it is a $1$-tensor on $\mathbb{R^n}$, simply by definition.
So in the equation \begin{equation} df(p)(v_p) = Df(p)(v), \end{equation} perhaps it might be useful to add another pair of brackts too see how things are being evaluated: \begin{equation} \left(df(p) \right)(v_p) = \left( Df(p) \right)(v). \end{equation} On the RHS, you are evaluating the linear transformation $Df(p): \mathbb{R^n} \to \mathbb{R}$ at the point $v \in \mathbb{R^n}$.